Problem: Evaluate the improper integral if it exists. $\int_{0}^{1}\dfrac{-4}{\sqrt[ 3]{x^2}}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-12$ (Choice B) B $3$ (Choice C) C $12$ (Choice D) D The improper integral diverges.
Explanation: First, let's rewrite the improper integral: $\int_{0}^{1}\dfrac{-4}{\sqrt[ 3]{x^2}}\,dx =\lim_{a\to0^+}\int_a^{1} \dfrac{-4}{\sqrt[ 3]{x^2}}\,dx$ We can now evaluate the integral: ∫ 1 0 1 x √ d x = lim a → 0 + ∫ 1 a − 4 x 2 − − √ 3 d x = lim a → 0 + ∫ 1 a − 4 x − 2 3 d x = − 4 lim a → 0 + ∫ 1 a x − 2 3 d x = − 4 lim a → 0 + [ 3 x 1 3 ] 1 a = − 4 lim a → 0 + ⎛ ⎝ 3 − 3 ( a ) 1 3 ⎞ ⎠ = − 4 ( lim a → 0 + 3 − lim a → 0 + 3 a 1 3 ) = − 4 ( 3 − 0 ) = − 12 \begin{aligned} \phantom{\int_0^{1}\dfrac1{\sqrt{x}}dx}&=\lim_{a\to0^+}\int_a^{1} \dfrac{-4}{\sqrt[ 3]{x^2}}\,dx\\ \\ \\ &=\lim_{a\to0^+}\int_a^{1}-4x\^{-\frac23}\,dx\\\\\\ &=-4\lim_{a\to0^+}\int_a^{1}x\^{-\frac23}\,dx\\\\\\ &=-4\lim_{a\to0^+}\Big[3x\^{\frac{1}{3}}\Big]_a^1\\ \\ \\ &=-4\lim_{a\to0^+}\left(3-3(a)\^{\frac{1}{3}}\right)\\ \\ \\ &=-4\Big(\lim_{a\to0^+}3-\lim_{a\to0^+}3a\^{\frac{1}{3}}\Big)\\ \\ &=-4(3-0)\\ \\ &=-12 \end{aligned} The answer: $\int_{0}^{1}\dfrac{-4}{\sqrt[ 3]{x^2}}\,dx =-12$